If f''x is continuous at x=0 and f''0=a then Ltx→02fx−3f2x+f4xx2=

Ltx→02fx−3f2x+f4xx2=Ltx→02f'x−6f'2x+4f'4x2x by LH rule=Ltx→02f"x−12f"2x+16f"4x2=f"0−6f"0+8f"0=3f"0=3a

If f''x is continuous at x=0 and f''0=a then Ltx→02fx−3f2x+f4xx2=

Ltx→02fx−3f2x+f4xx2=Ltx→02f'x−6f'2x+4f'4x2x by LH rule=Ltx→02f"x−12f"2x+16f"4x2=f"0−6f"0+8f"0=3f"0=3a

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If $f'' (x)$ is continuous at x=0 and $f'' (0) = a$ then $\underset{x \to 0}{L t} \frac{2 f (x) - 3 f (2 x) + f (4 x)}{x^{2}} =$

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answer is C.

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Detailed Solution

$\begin{matrix} \underset{x \to 0}{L t} \frac{2 f (x) - 3 f (2 x) + f (4 x)}{x^{2}} = \underset{x \to 0}{L t} \frac{2 f' (x) - 6 f' (2 x) + 4 f' (4 x)}{2 x} b y L H r u l e \\ = \underset{x \to 0}{L t} \frac{2 f " (x) - 12 f " (2 x) + 16 f " (4 x)}{2} \\ = f " (0) - 6 f " (0) + 8 f " (0) \\ = 3 f " (0) \\ = 3 a \end{matrix}$